If $X$ is $k$-regular and $\theta_r\ne k$, then $\one^TU_r=0$. Hence the vertices of $X$ form a 1-design.
\[ U = \begin{pmatrix} -0.3873& 0.0& 0.0\\ -0.2887& -0.1706& 0.1938\\ -0.1291& -0.0228& 0.3644\\ -0.1291& 0.2391& 0.2760\\ 0.1291& 0.3042& 0.2020\\ 0.2887& 0.0826& 0.2446\\ 0.1291& -0.1195& 0.3450\\ 0.1291& -0.3270& 0.1625\\ -0.1291& -0.3586& 0.0690\\ & \vdots& \end{pmatrix} \]
Let $U$ be an $m\times m$ matrix whose columns form an orthonormal basis for the $\tau$-eigenspace. Then there is a vector $h$ in $\re^m$ such that \[ x - \frac{|S|}{n}\one = Uh. \]
It follows that we can partition the $\tau$-eigenpolytope into a parallel pair of faces.
Suppose $Uh$ is the shifted characteristic vector of a coclique, and denote its entries by $\alpha$ and $\beta$. Then the function \[ (h^Tx-\alpha)(h^Tx-\beta) \] is a quadratic polynomial that vanishes on the image of each vertex of $X$.
The Veronese map $\ver_k$ is a mapping from a vector space of dimension $m$ to a vector space of dimension $\binom{m+k}{k}$ such that the entries of $\ver_k(x)$ are all monomials of degree at most $k$ in the entries of $x$.
An example: if \[ x = \begin{pmatrix}a&b\end{pmatrix} \] then \[ \ver_2(x)= \begin{pmatrix}1&a&b&a^2&ab&b^2\end{pmatrix} \]
If $U$ is a matrix, then $\ver_k(U)$ is the matrix we get by applying $\ver_k$ to each row.
Lemma: There is a quadratic polynomial that vanishes on each row of $U$ if and only if the columns of $\ver_2(U)$ are linearly dependent.
Find graphs where we can use the rank of the Veronese of an eigenspace to rule out the existence of ratio-tight cocliques.
A subset $S$ of the unit sphere in $\re^d$ has strength $t$ if the average value over $S$ of any polynomial of degree at most $t$ is equal to its average over the entire sphere. We say $S$ is a $t$-design if its strength is at least $t$.
Thus $S$ is a 1-design if and only if the sum of its elements is zero.If $X$ is $k$-regular and $\theta_r\ne k$, then $\one^TU_r=0$. Hence the vertices of $X$ form a 1-design.
If $h$, $\alpha$, $\beta$ are such that $(h^Tx-\alpha)(h^Tx-\beta)$ is zero on all vertices of $X$, then \[ (h^Tx-\alpha)(h^Tx-\beta) \ell^Tx \] is zero on all vertices.
If the vertices of $X$ form a 3-design, then it follows that $(h^Tx-\alpha)(h^Tx-\beta)$ is orthogonal to all linear functions.
In particular it must be orthogonal to $h^Tx$.
If the $\tau$-eigenspace is a 3-design and there is a ratio-tight coclique, then $X$ is bipartite and the coclique is a color class.